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Molarity Definition

A way of indicating the concentration of a solution. The amount of solute expressed in terms of moles. There are two main types:

(1) Weight molar concentration or heavy mold concentration. The concentration expressed as the number of moles of solute dissolved per 1000 grams of solvent.

Commonly used m as a symbol. For example, a solution of 1 gram of sulfuric acid (98.08 g) dissolved in 1000 g of water at a concentration of 1 m.

(2) Volume molar concentration or mold concentration. The concentration expressed as the number of moles of the solute contained in 1 liter of the solution.

Commonly used M as a symbol. For example, when 1 gram of sulfuric acid (98.08 g) is dissolved in water to form a 1 liter solution, the concentration is 1 M.

Calculation of Molar Concentration

The solute is expressed in a specific weight unit, ie, “molecule”. The molecular weight is expressed in grams, which is called a mole. The number of moles refers to a few moles.

For example, sodium hydroxide (NaOH) has a molecular weight of 40, so 40 grams of sodium hydroxide is 1 mole of sodium hydroxide, and the number of moles is equal to 1. 80 grams of sodium hydroxide is 2 moles, and the number of moles is equal to 2.

To convert a known weight into a mole number, simply divide the grams of the substance by the molecular weight of the substance: the number of grams = the grams of the substance / the molecular weight of the substance. For example, 120 grams of NaOH = 140/30 = 3 moles.

The molar concentration is expressed in terms of the number of moles of solute contained in 1 liter of solution and is represented by the letter “M” (upper case).

For example, in a 1 liter NaOH solution, if it contains 40 grams of NaOH (ie, 1 mole of NaOH), then the solution is 1 mole, or 1M; if 2 liters of solution contains 8 grams of NaOH (ie 0.2 moles), this The solution concentration is 0.1M, and so on.

According to the definition of the molar concentration, it can be obtained: M = number of solute moles/solution liter number, shift term: M × solution liters = solute mole number.

The molar concentration is less applicable in the volume analysis than the complex titration application.

However, the molecule is one of the basic concepts in chemistry, so understanding its meaning and proficiency in the calculation of the concentration of molecules is helpful in understanding the calculation of the concentration expressed in other units.

Volumetric molar concentration and weight molar concentration

Percent concentration solutions are easy to formulate, but their use is limited. The 5% sodium chloride solution and the 5% glucose solution contain the same weight of solute but contain different solute mole numbers.

Because the chemical amounts of sodium chloride and glucose are different. Mercury is very useful chemically. Therefore, there is a method based on the concentration of the ketone, called the molar concentration.

The symbol of the molar concentration is M, which is defined as the number of grams of solute contained per liter (1000 ml) of the solution: volume molar concentration (M) = number of moles of solute per liter of solution.

Therefore, a solution of 1 vol. (1 M) is one gram of solute per liter of solution. A solution of 2 vol. (2 M) is a 2 gram solute per liter of solution, and so on. A solution of 1/2 vol. (0.5 M) concentration contains less than 1 gram of sol, only 0.5 gram of solute per liter of solution.

Example: 0.15MNaCl

0.15MNaCl=0.15 moles NaCl/liter solution

0.15 mol NaCl = 0.15 mol X58 g / mol = 8.7 g

0.15MNaCl = 8.7 grams of NaCl per liter of solution

First, the meaning of the symbol 0.15M is written. Then, since the chemical formula of sodium chloride is 58 amu, the mass of one molecule is 58 g. By multiplying the number of molecules by 1 mole of mass, a mass of 0.15 moles can be obtained. Thus, a 0.15 M sodium chloride solution per liter contains 8.7 grams of sodium chloride.

One liter of 0.9% NaCI solution contained 9 grams of sodium chloride. Thus, 0.15 M NaCI and 0.9% NaCI are two different representations of approximately the same concentration.

One is expressed in grams per liter and the other is expressed in grams per 100 milliliters. However, the sodium chloride content of the solution is approximately the same. Therefore, both can be formulated in the same way.

If a 1 liter (1000 ml) solution is required, the solute weight must be adjusted to accommodate the desired solution volume.

For example, to formulate 2 liters of 0.15 M NaCl solution, 1 liter of twice the amount of sodium chloride used is required: 2 x 9 grams, or 18 grams. To formulate less than 1 liter of solution, such as 500 milliliters (0.5 liters), 0.5 x 9 grams, or 4.5 grams of sodium chloride, is required.

These solutions contained 9 grams in 1 liter, 18 grams in 2 liters, and 4.5 liters in 0.5 liters at exactly the same concentration. The ratio of the number of moles of the solute to the number of liters of the solution is the same, ie 0.15.

A concentration representation similar to the volumetric molar concentration is the gram mole concentration with a sign of 91.

It is defined as the number of moles of solute per kilogram (kg) of solvent. In the volumetric molar concentration is the ratio of the number of moles to the total volume of the solution, and the weight concentration is the ratio of the number of moles of the solute to the mass of the solvent.

For example, a 0.15-gram molar sodium chloride solution is composed of 0.15 gram (9 grams) of sodium chloride dissolved in 1000 grams of water. For a dilute aqueous solution, since the volume of 1000 g of water is 1 liter, the molar concentration and the molar concentration are approximately equal.

Molarity calculations are a “must learn” in general or organic labs.  Before going into Molarity calculations, it is important that you brush up on your stoichiometry. Chemistry Molarity Calculations

Molarity Calculations

We must also understand what concentration is.  I like the definition which my chemistry book gives.

It states Concentration is a measure of the quantity of solute dissolved in a given quantity of solutions.

This definition was taken from Silberberg’s Fourth Edition Chemistry book – Chemistry, The Molecular Nature of Matter and Change.  To sum up, what this means, this means that if you add more of the solute in the solution, you will have a higher concentration.  If this doesn’t make sense right now, don’t worry!

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Now let’s talk about what Molarity is.  Molarity is just another unit, abbreviated M (not to be mistaken for moles).  The unit for Molarity is Moles of solute per Liter of the solution or (mol/L).

Let’s dig into the main concept in this chapter… How to solve Molarity calculations.

In most all Molarity problems, you will be given three out of four variables and will solve for the fourth.  The four variables are:

The concentration of A

The volume of A

The concentration of B

The volume of B

A general formula for solving Molarity problems is the following:

Mdil x Vdil = number of moles = Mconc x Vconc

Don’t think you need to solve every Molarity problem using the above equation, you can also do them by unit conversions. Chemistry Molarity Calculations What to Read Next:-

Let me do a sample Molarity problem so you can see how the above equation works.

Problem 1

Problem: If 5.0 L of 3.50 M acid is to makes a 0.500 M solution, what is the final volume of the solution when diluted to 0.500 M?  Assume this is a 1 to 1 mole ratio.

Plan: In order to solve this problem, we will use the above equation Mdil x Vdil = number of moles = Mconc x Vconc. We know we have the concentration (molarity) of A, the volume of A and the concentration of B.  Therefore, let’s solve for the volume of B.

(Mdil x Vdil)/Mconc =  Vconc

Answer: By filling in the above algebra equation we get:

Discussion: A good way to determine if you are performing the problem correctly, is to check if your units cancel to give what the problem is asking for.

Here both the molarities cancel out and give liters.  Because the problem states that this was a 1 to 1 mole ratio,

the “=number of moles” can be ignored.  This is the case in about 60% of Molarity problems.

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